Problem B

Back to High School Physics

Input: standard input

Output: standard output

A particle has initial velocity and constant acceleration. If its velocity after certain time is v then what will its displacement be in twice of that time?

Input

The input will contain two integers in each line. Each line makes one set of input. These two integers denote the value of v (-100 <= v <= 100) and t(0<=t<= 200) ( t means at the time the particle gains that velocity)

Output

For each line of input print a single integer in one line denoting the displacement in double of that time.

Sample Input

0 0
5 12

Sample Output

0
120

___________________________________________________________________________________
Shahriar Manzoor

 #include <stdio.h>
int main()
{
int v, t;
for (;scanf("%d%d", &v, &t) != EOF;)
if (v <= && v >= - || t >= && t <= )
printf("%d\n", v * t * );
return ;
}

报告:这道题没有什么好说的,本来已经准备好跟第一题一样大战三百回合的,结果一轮AC了。

问题:无

解决:无

UVa OJ 10071的更多相关文章

  1. uva oj 567 - Risk(Floyd算法)

    /* 一张有20个顶点的图上. 依次输入每个点与哪些点直接相连. 并且多次询问两点间,最短需要经过几条路才能从一点到达另一点. bfs 水过 */ #include<iostream> # ...

  2. UVa OJ 194 - Triangle (三角形)

    Time limit: 30.000 seconds限时30.000秒 Problem问题 A triangle is a basic shape of planar geometry. It con ...

  3. UVa OJ 175 - Keywords (关键字)

    Time limit: 3.000 seconds限时3.000秒 Problem问题 Many researchers are faced with an ever increasing numbe ...

  4. UVa OJ 197 - Cube (立方体)

    Time limit: 30.000 seconds限时30.000秒 Problem问题 There was once a 3 by 3 by 3 cube built of 27 smaller ...

  5. UVa OJ 180 - Eeny Meeny

    Time limit: 3.000 seconds限时3.000秒 Problem问题 In darkest <name of continent/island deleted to preve ...

  6. UVa OJ 140 - Bandwidth (带宽)

    Time limit: 3.000 seconds限时3.000秒 Problem问题 Given a graph (V,E) where V is a set of nodes and E is a ...

  7. 548 - Tree (UVa OJ)

    Tree You are to determine the value of the leaf node in a given binary tree that is the terminal nod ...

  8. UVa OJ 10300

    Problem A Ecological Premium Input: standard input Output: standard output Time Limit: 1 second Memo ...

  9. UVa OJ 10055

    Problem A Hashmat the brave warrior Input: standard input Output: standard output Hashmat is a brave ...

随机推荐

  1. 【转】Java中只有按值传递,没有按引用传递!

    原文链接:http://guhanjie.iteye.com/blog/1683637 今天,我在一本面试书上看到了关于java的一个参数传递的问题: 写道 java中对象作为参数传递给一个方法,到底 ...

  2. JNLP + Applet + Bouncy Castle

    http://stackoverflow.com/questions/4275005/jnlp-applet-bouncy-castle ——————————————————————————————— ...

  3. OperateParticleWithCodes

    [OperateParticleWithCodes] Listing 6-6 shows how you might configure an emitter’s scale property. Th ...

  4. getGuid()

    function GetGUID: string;var  LTep: TGUID;begin  CreateGUID(LTep);  Result := GUIDToString(LTep);  R ...

  5. POJ-2785 4 Values whose Sum is 0(折半枚举 sort + 二分)

    题目链接:http://poj.org/problem?id=2785 题意是给你4个数列.要从每个数列中各取一个数,使得四个数的sum为0,求出这样的组合的情况个数. 其中一个数列有多个相同的数字时 ...

  6. Odoo的Domain (一)

    Odoo 的Domain:多个条件的列表. 条件:(字段名,操作符,值)三元式(列表或者元组) 字段名:当前模型的字段或者是通过点操作符访问的Many2one/Many2Many对象,当是Many2M ...

  7. InitializingBean和init-method

    [spring的InitializingBean的 afterPropertiesSet 方法 和 init-method配置的 区别联系] InitializingBean Spring的Initi ...

  8. HTML5 Web Speech API 结合Ext实现浏览器语音识别以及输入

    简介      Web Speech API是HTML5技术中的一个提供原生语音识别技术的API,Google Chrome在25版之后开始支持Web Speech API,Google也提供了一个 ...

  9. 切取图片的一部分(利用CCRenderTexture)

    转自:http://blog.csdn.net/wcjwdq/article/details/37932769 显示图片时,在项目中经常会用只读取图片的一部分,而不是全部. 错误方式:很多人这时候会采 ...

  10. uva301 - Transportation

      Transportation Ruratania is just entering capitalism and is establishing new enterprising activiti ...